360R-06 Design of Slabs-on-Ground


partially in the 30 ft direction before the adjacent strip is



tải về 2.35 Mb.
Chế độ xem pdf
trang103/107
Chuyển đổi dữ liệu11.08.2022
Kích2.35 Mb.
#52863
1   ...   99   100   101   102   103   104   105   106   107
Design of Slabs-on-Ground


partially in the 30 ft direction before the adjacent strip is
placed. Final stress will tie all strips together on the end.
When calculating the force to overcome the subgrade
friction, the total width of all strips is to be considered (12 x
30 = 360 ft).
Case 2: A section of 200 ft is placed first, stressed partially,
and then the other section of 160 ft is placed and stressed.
When calculating the force to overcome the subgrade
friction, use the following criteria:
Placement 1: Formula
W
slab
× × 
μ
A
Placement 2:
B
The tendons in Placement 1 have to overcome maximum
friction based on 180 ft length at the critical section at the
center of the combined length (dashed line).
The tendons in Placement 2 have to overcome maximum
friction based on 160 ft length at the critical section at the
joint between Placement 1 and 2 (pulling Placement 2
toward Placement 1).
A4.3—Design example: Equivalent tensile
stress design
Determine the reduction in slab thickness of a 6 in. thick
unreinforced slab if post-tensioning is used.
Assume a modulus of rupture of 9
with a safety factor
of 2 was used to design the unreinforced 6 in. thick slab.
Then, the allowable tension stress for 4000 psi concrete
would be
= 285 psi
If the P-T force will provide an effective residual compression
of 150 psi (selected for this example) with the tendons in the
center of the slab, then the allowable tensile stress due to the
bending moments is 150 psi + 285 psi = 435 psi.
The moment capacity of the slab is given by
L
2
---
6 in.
12 in./ft
-------------------
150 lb/ft
3
500 ft
2
--------------
0.5
×
×
×
P
e
f
p
WH
P
r
+
-------------------------
26,000 lb
250 psi
12 in.
6 in.
9375 lb/ft
×
×
×
------------------------------------------------------------------------------------
P
h
2
-----
h
3
( )
log
1.6a
2
h
2
+
0.675h
)

f
c

L
2
---
L
2
---
360
2
---------
180 ft
=
=
L
2
---
160 ft
=
f
c

9 4000 psi
2
----------------------------



tải về 2.35 Mb.

Chia sẻ với bạn bè của bạn:
1   ...   99   100   101   102   103   104   105   106   107




Cơ sở dữ liệu được bảo vệ bởi bản quyền ©hocday.com 2024
được sử dụng cho việc quản lý

    Quê hương